Two wavelengths of sodium light \( 590\,\mathrm{nm} \) and \( 595\,\mathrm{nm} \) are used in a diffraction experiment having a single slit of aperture \( 10^{-4}\,\mathrm{m} \). The distance between the slit and the screen is \( 1.5\,\mathrm{m} \). The separation between the positions of the first maxima of the diffraction pattern obtained in the two cases is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \sim 0.11\,\mathrm{mm} \)
In single-slit diffraction the central maximum is at the same place for both wavelengths, so “first maxima” means the first secondary maxima, which lie approximately midway between the first and second minima:
\( \displaystyle \theta_1 \approx \frac{3\lambda}{2a} \Rightarrow y_1 = \frac{3\lambda D}{2a} \)
For the two wavelengths the separation is
\( \displaystyle \Delta y = \frac{3D}{2a}\,(\lambda_2-\lambda_1) \)
\( \lambda_2-\lambda_1 = 595-590 = 5\,\mathrm{nm} = 5\times10^{-9}\,\mathrm{m} \)
\( D = 1.5\,\mathrm{m} \)
\( a = 10^{-4}\,\mathrm{m} \)
\( \displaystyle \Delta y = \frac{3\times1.5\times5\times10^{-9}}{2\times10^{-4}} = 1.125\times10^{-4}\,\mathrm{m} \)
\( \Delta y \approx 0.11\,\mathrm{mm} \)